Lets S={z in C:|z-1|=1 and (sqrt{2}-1)(z+bar{ | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Let $Z=x+i y$

Then $(x-1)^2+y^2=1 \rightarrow(1)$

$(\sqrt{2}-1)(2 x)-i(2 i y)=2 \sqrt{2}$

$\quad \Rightarrow(\sqrt{2}-1) x+y=\sqrt{2} \righta

Vedclass · Accepted Answer

$2$

Vedclass · Answer

Let $Z=x+i y$

Then $(x-1)^2+y^2=1 \rightarrow(1)$

$(\sqrt{2}-1)(2 x)-i(2 i y)=2 \sqrt{2}$

$\quad \Rightarrow(\sqrt{2}-1) x+y=\sqrt{2} \rightarrow(2)$

Solving $(1) and (2)$ we get

Either $\mathrm{x}=1$ or $x=\frac{1}{2-\sqrt{2}} \rightarrow$ $(3)$

On solving $(3)$ with $(2)$ we get

For $x=1 \Rightarrow y=1 \Rightarrow Z_2=1+i$  and  for

$x=\frac{1}{2-\sqrt{2}} \Rightarrow y=\sqrt{2}-\frac{1}{\sqrt{2}} \Rightarrow Z_1=\left(1+\frac{1}{\sqrt{2}}\right)+\frac{i}{\sqrt{2}}$

Now

$\left|\sqrt{2} z_1-z_2\right|^2$

$=\left|\left(\frac{1}{\sqrt{2}}+1\right) \sqrt{2}+i-(1+i)\right|^2$

$=(\sqrt{2})^2$

$=2$

List $I$	List $II$
$P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j=1$	$1.$ True
$Q.$ There exists a $k \in\{1,2, \ldots ., 9\}$ such that $z_{1 .} . z=z_k$ has no solution $z$ in the set of complex numbers.	$2.$ False
$R.$ $\frac{\left\|1-z_1\right\|\left\|1-z_2\right\| \ldots . .\left\|1-z_9\right\|}{10}$ equals	$3.$ $1$
$S.$ $1-\sum_{k=1}^9 \cos \left(\frac{2 k \pi}{10}\right)$ equals	$4.$ $2$

Lets $S=\{z \in C:|z-1|=1$ and $(\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2 \sqrt{2}\}$. Let $\mathrm{z}_1, \mathrm{z}_2$ $\in S$ be such that $\left|z_1\right|=\max _{z \in S}|z|$ and $\left|z_2\right|=\min _{z \in S}|z|$. Then $\left|\sqrt{2} z_1-z_2\right|^2$ equals :

Similar Questions

Amplitude of $\left( {\frac{{1 - i}}{{1 + i}}} \right)$ is

The inequality $|z - 4|\, < \,|\,z - 2|$represents the region given by

Find the modulus and the argument of the complex number $z=-1-i \sqrt{3}$.

The moduli of two complex numbers are less than unity, then the modulus of the sum of these complex numbers